Sum of first n natural numbers

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Since \(5050=\frac{100\times101}{2}\), we get \(n=100\). Therefore, \(S_{n-10}=S_{90}=\frac{90\times91}{2}=4095\). Option 4186 is not correct because it is the sum of the first 91 natural numbers. Exam tip: Recognise 5050 as \(S_{100}\), then reduce the subscript by 10.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{45}=\frac{45\times46}{2}=1035\) and \(S_{30}=\frac{30\times31}{2}=465\). Therefore, \(S_{45}:S_{30}=1035:465=69:31\). The option \(45:30\) is only the ratio of the numbers of terms, not the ratio of their sums. Exam tip: For ratios of such sums, apply \(\frac{n(n+1)}{2}\) first and then cancel common factors.

Here, \(S_n\) is the sum of the first \(n\) natural numbers. In \(S_n-S_{n-2}\), the first \(n-2\) terms cancel, leaving \((n-1)+n=2n-1\). Thus, \(2n-1=101\) gives \(n=51\). Therefore, \(S_{51}=\frac{51\times52}{2}=1326\). Option 1275 is \(S_{50}\), so it is close but not correct. Exam tip: In differences of partial sums, cancel the common initial terms and write the remaining terms.

There are \(45-15+1=31\) terms from 15 to 45. The sum of the first and last terms is \(15+45=60\), so the sum is \(\frac{31\times 60}{2}=930\). Although 960 is close, it does not follow from the correct number of terms and average of this sequence. Exam tip: When both endpoints are included, always use \(+1\) while counting the terms.

\(S_{n+1}=1+2+\cdots+n+(n+1)\) and \(S_{n-1}=1+2+\cdots+(n-1)\). On subtracting, the common terms cancel, leaving \(n+(n+1)=2n+1\). Hence, the correct answer is \(2n+1\). \(n+1\) is only the newly added last term; the term \(n\) must also be included. Exam tip: Expand the two sums and identify the terms that cancel.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{72}=\frac{72\times73}{2}=2628\) and \(S_{48}=\frac{48\times49}{2}=1176\). Hence, the difference is \(2628-1176=1452\). It is also the sum of the numbers from 49 to 72; an option such as 1462 results from an arithmetic error. Exam tip: for such questions, write \(S_{72}-S_{48}\) directly.

Using \(S_n=\frac{n(n+1)}{2}\), the sum is odd when n leaves remainder 1 or 2 on division by 4. Remainders 0 and 3 give an even sum. Exam tip: memorise this mod-4 pattern.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=1081\), we get \(n=46\). Hence, \(S_{n+4}-S_n\) is the sum of the next four terms: \(47+48+49+50=194\). Although 198 is close, it is not the correct sum of these four consecutive numbers. Exam tip: Rewrite \(S_{n+k}-S_n\) as the sum of terms from \(n+1\) to \(n+k\).

There are \(80-40+1=41\) terms from 40 to 80. Therefore, the sum is \(\frac{41(40+80)}{2}=41\times 60=2460\). Option 2420 may result from an error in counting the terms or finding the average. Exam tip: the number of integers from a first value to a last value is always \(\text{last}-\text{first}+1\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(S_n-S_{50}=1575\) and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, \(S_n=1575+1275=2850\). Now \(\frac{n(n+1)}{2}=2850\), so \(n(n+1)=5700=75\times76\). Hence, \(n=75\). The nearby option 80 does not give a difference of 1575 from \(S_{50}\). Exam tip: In such questions, first add the known \(S_k\) to find \(S_n\).

The sum of the first n natural numbers is n(n+1)/2. One of n and n+1 is even; after dividing by 2, the remaining factor must also be even. Thus the sum is even when n ≡ 0 or 3 (mod 4). Exam tip: check n modulo 4 directly.

If \(T=\frac{n(n+1)}{2}\), then \(8T+1=4n(n+1)+1=(2n+1)^2\), which is an odd perfect square. The converse is also true. \(4T+1\) has no such guaranteed form. Exam tip: test \(8T+1\).

Let \(S_n\) be the sum of the first \(n\) natural numbers. In \(S_{n+2}-S_n\), the common initial terms cancel, leaving \((n+1)+(n+2)=2n+3\). Thus, \(2n+3=155\), so \(2n=152\) and \(n=76\). If \(n=77\), the difference would be \(157\), not \(155\). Exam tip: In differences of sums, cancel the common terms and add only the remaining terms.

The sum of the first 120 natural numbers is \(\frac{120\times121}{2}=7260\), and the sum of the first 80 natural numbers is \(\frac{80\times81}{2}=3240\). Therefore, the required result is \(7260-3240=4020\). This is also the sum of the numbers from 81 to 120. An option such as 4040 can result from an arithmetic error in subtraction. Exam tip: use \(S_n=\frac{n(n+1)}{2}\) and subtract the two sums carefully.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{90}=\frac{90\times91}{2}=4095\) and \(S_{60}=\frac{60\times61}{2}=1830\). Therefore, \(4095:1830=273:122\). The ratio \(3:2\) is only the ratio of 90 to 60, not of their sums. Exam tip: When comparing such sums, include the \(n+1\) factor in \(n(n+1)/2\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=3655\), we get \(n=85\), since \(\frac{85\times86}{2}=3655\). Therefore, \(S_n-S_{n-4}=S_{85}-S_{81}\), which is the sum of the last four terms: \(82+83+84+85=334\). The value 330 would result from adding \(81+82+83+84\), which is not the required set of final four terms. Exam tip: \(S_n-S_{n-r}\) equals the sum of the last r terms up to n.

The sum is \(S=\frac{n(n+1)}{2}\). Hence \(8S+1=4n(n+1)+1=(2n+1)^2\), which is an odd perfect square. Exam tip: use this identity to test whether a number is triangular.

There are \(75-25+1=51\) terms from 25 to 75. These numbers form an arithmetic progression, so their sum is \(\frac{51}{2}(25+75)=\frac{51}{2}\times100=2550\). The option 2500 results from not using the number of terms or the average correctly. Exam tip: for an inclusive range from \(a\) to \(b\), always use \(b-a+1\) terms.

To obtain the sum of the first \(n\) natural numbers from \(S_{n-1}\), add the new term \(n\). Hence \(S_n=S_{n-1}+n\). Option B adds the previous term instead. Exam tip: compare consecutive partial sums to identify the added term.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(S_{84}=\frac{84\times85}{2}=3570\). Half of this sum is \(\frac{3570}{2}=1785\), so option B is correct. A nearby value such as \(1815\) may look plausible, but it is not half of the required sum. Exam tip: first find the total using \(n(n+1)/2\), then apply the fraction asked for—in this case, one-half.